6.9E: Mazoezi ya Sehemu ya 6.9

1) [T] Kupata maneno kwa\(\cosh x+\sinh x\) na\(\cosh x−\sinh x.\) kutumia calculator kwa grafu kazi hizi na kuhakikisha kujieleza yako ni sahihi.

Jibu

\(e^x\)na\(e^{−x}\)

2) Kutoka kwa ufafanuzi wa\(\cosh(x)\) na\(\sinh(x)\), pata antiderivatives yao.

3) Onyesha hilo\(\cosh(x)\) na\(\sinh(x)\) kukidhi\( y''=y\).

Jibu

Majibu inaweza kutofautiana

4) Tumia utawala wa quotient ili uhakikishe kuwa\(\dfrac{d}{dx}\big(\tanh(x)\big)=\text{sech}^2(x).\)

5) Pata\(\cosh^2(x)+\sinh^2(x)=\cosh(2x)\) kutoka kwa ufafanuzi.

Jibu

Majibu inaweza kutofautiana

6) Chukua derivative ya kujieleza uliopita ili kupata maelezo ya\(\sinh(2x)\).

7) Thibitisha\(\sinh(x+y)=\sinh(x)\cosh(y)+\cosh(x)\sinh(y)\) kwa kubadilisha maneno kwa maonyesho.

Jibu

Majibu inaweza kutofautiana

8) Chukua derivative ya kujieleza uliopita ili kupata maelezo\(\cosh(x+y).\)

Katika mazoezi 9 - 18, pata derivatives ya kazi zilizopewa na grafu pamoja na kazi ili kuhakikisha jibu lako ni sahihi.

9) [T]\(\cosh(3x+1)\)

Jibu

\(3\sinh(3x+1)\)

10) [T]\(\sinh(x^2)\)

11) [T]\(\dfrac{1}{\cosh(x)}\)

Jibu

\(−\tanh(x)\text{sech}(x)\)

12) [T]\(\sinh(\ln(x))\)

13) [T]\(\cosh^2(x)+\sinh^2(x)\)

Jibu

\(4\cosh(x)\sinh(x)\)

14) [T]\(\cosh^2(x)−\sinh^2(x)\)

15) [T]\(\tanh(\sqrt{x^2+1})\)

Jibu

\(\dfrac{x\text{sech}^2(\sqrt{x^2+1})}{\sqrt{x^2+1}}\)

16) [T]\(\dfrac{1+\tanh(x)}{1−\tanh(x)}\)

17) [T]\(\sinh^6(x)\)

Jibu

\(6\sinh^5(x)\cosh(x)\)

18) [T]\(\ln(\text{sech}(x)+\tanh(x))\)

Katika mazoezi 19 - 28, pata antiderivatives kwa kazi zilizopewa.

19)\(\cosh(2x+1)\)

Jibu

\(\frac{1}{2}\sinh(2x+1)+C\)

20)\(\tanh(3x+2)\)

21)\(x\cosh(x^2)\)

Jibu

\(\frac{1}{2}\sinh^2(x^2)+C\)

22)\(3x^3\tanh(x^4)\)

23)\(\cosh^2(x)\sinh(x)\)

Jibu

\(\frac{1}{3}\cosh^3(x)+C\)

24)\(\tanh^2(x)\text{sech}^2(x)\)

25)\(\dfrac{\sinh(x)}{1+\cosh(x)}\)

Jibu

\(\ln(1+\cosh(x))+C\)

26)\(\coth(x)\)

27)\(\cosh(x)+\sinh(x)\)

Jibu

\(\cosh(x)+\sinh(x)+C\)

28)\((\cosh(x)+\sinh(x))^n\)

Katika mazoezi 29 - 35, tafuta derivatives kwa kazi.

29)\(\tanh^{−1}(4x)\)

Jibu

\(\dfrac{4}{1−16x^2}\)

30)\(\sinh^{−1}(x^2)\)

31)\(\sinh^{−1}(\cosh(x))\)

Jibu

\(\dfrac{\sinh(x)}{\sqrt{\cosh^2(x)+1}}\)

32)\(\cosh^{−1}(x^3)\)

33)\(\tanh^{−1}(\cos(x))\)

Jibu

\(−\csc(x)\)

34)\(e^{\sinh^{−1}(x)}\)

35)\(\ln(\tanh^{−1}(x))\)

Jibu

\(−\dfrac{1}{(x^2−1)\tanh^{−1}(x)}\)

Katika mazoezi 36 - 42, tafuta antiderivatives kwa kazi.

36)\(\displaystyle ∫\frac{dx}{4−x^2}\)

37)\(\displaystyle ∫\frac{dx}{a^2−x^2}\)

Jibu

\(\dfrac{1}{a}\tanh^{−1}\left(\dfrac{x}{a}\right)+C\)

38)\(\displaystyle ∫\frac{dx}{\sqrt{x^2+1}}\)

39)\(\displaystyle ∫\frac{xdx}{\sqrt{x^2+1}}\)

Jibu

\(\sqrt{x^2+1}+C\)

40)\(\displaystyle ∫−\frac{dx}{x\sqrt{1−x^2}}\)

41)\(\displaystyle ∫\frac{e^x}{\sqrt{e^{2x}−1}}\)

Jibu

\(\cosh^{−1}(e^x)+C\)

42)\(\displaystyle ∫−\frac{2x}{x^4−1}\)

Katika mazoezi 43 - 45, tumia ukweli kwamba mwili unaoanguka na msuguano sawa na kasi ya mraba hutii equation\(\dfrac{dv}{dt}=g−v^2\).

43) Onyesha kwamba\(v(t)=\sqrt{g}\tanh(\sqrt{g}t)\) satisfies equation hii.

Jibu

Majibu inaweza kutofautiana

44) Pata usemi uliopita\(v(t)\) kwa kuunganisha\(\dfrac{dv}{g−v^2}=dt\).

45) [T] Tathmini jinsi mbali mwili imeshuka katika\(12\) sekunde kwa kutafuta eneo chini ya Curve ya\(v(t)\).

Jibu

\(37.30\)

Katika mazoezi 46 - 48, tumia hali hii: Cable kunyongwa chini ya uzito wake ina mteremko\(S=\dfrac{dy}{dx}\) unaotimiza\(\dfrac{dS}{dx}=c\sqrt{1+S^2}\). Mara kwa mara\(c\) ni uwiano wa wiani wa cable kwa mvutano.

46) Onyesha kwamba\(S=\sinh(cx)\) satisfies equation hii.

47)\(\dfrac{dy}{dx}=\sinh(cx)\) Unganisha ili kupata urefu cable\(y(x)\) kama\(y(0)=1/c\).

Jibu

\(y=\frac{1}{c}\cosh(cx)\)

48) Mchoro cable na kuamua jinsi mbali chini sags katika\(x=0\).

Katika mazoezi 49 - 52, tatua kila tatizo.

49) [T] mlolongo hangs kutoka posts mbili\(2\) m mbali na kuunda catenary ilivyoelezwa na equation\(y=2\cosh(x/2)−1\). Pata mteremko wa catenary kwenye chapisho la uzio wa kushoto.

Jibu

\(−0.521095\)

50) [T] mlolongo hangs kutoka posts mbili mita nne mbali na kuunda catenary ilivyoelezwa na equation\(y=4\cosh(x/4)−3.\) Kupata urefu jumla ya catenary (safu urefu).

51) [T] Mstari wa nguvu ya juu-voltage ni catenary iliyoelezwa na\(y=10\cosh(x/10)\). Pata uwiano wa eneo chini ya catenary kwa urefu wake wa arc. Unaona nini?

Jibu

\(10\)

52) Mstari wa simu ni catenary iliyoelezwa na\(y=a\cosh(x/a).\) Pata uwiano wa eneo chini ya catenary kwa urefu wake wa arc. Je, hii inathibitisha jibu lako kwa swali la awali?

53) Thibitisha formula kwa derivative ya\(y=\sinh^{−1}(x)\) kwa kutofautisha\(x=\sinh(y).\)

(Kidokezo: Tumia utambulisho wa trigonometric ya hyperbolic.)

54) Thibitisha formula kwa derivative ya\(y=\cosh^{−1}(x)\) kwa kutofautisha\(x=\cosh(y).\)

(Kidokezo: Tumia utambulisho wa trigonometric ya hyperbolic.)

55) Thibitisha formula kwa derivative ya\(y=\text{sech}^{−1}(x)\) kwa kutofautisha\(x=\text{sech}(y).\)

(Kidokezo: Tumia utambulisho wa trigonometric ya hyperbolic.)

56) Thibitisha kwamba\(\cosh(x)+\sinh(x))^n=\cosh(nx)+\sinh(nx).\)

57) Thibitisha usemi\(x=\sinh(y)=\dfrac{e^y−e^{−y}}{2}\) kwa\(\sinh^{−1}(x).\) Kuzidisha\(2e^y\) na kutatua kwa\(y\). Je! Maneno yako yanafanana na kitabu?

58) Thibitisha usemi\(x=\cosh(y)=\dfrac{e^y+e^{−y}}{2}\) kwa\(\cosh^{−1}(x).\) Kuzidisha\(2e^y\) na kutatua kwa\(y\). Je! Maneno yako yanafanana na kitabu?

Wachangiaji

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